Problem: You are given N items.
The value of the i-th item (1 \leq i \leq N) is v_i.
Your have to select at least A and at most B of these items.
Under this condition, find the maximum possible arithmetic mean of the values of selected items.
Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

* 1 \leq N \leq 50
* 1 \leq A,B \leq N
* 1 \leq v_i \leq 10^{15}
* Each v_i is an integer.

Input

The input is given from Standard Input in the following format:


N A B
v_1
v_2
...
v_N


Output

Print two lines.
The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.
The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Examples

Input

5 2 2
1 2 3 4 5


Output

4.500000
1


Input

4 2 3
10 20 10 10


Output

15.000000
3


Input

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996


Output

1000000000000000.000000
1

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

/**
 * 解説どおりに作成したソースコード
 */
public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			int a = scanner.nextInt();
			int b = scanner.nextInt();
			// m_C_nの配列を計算
			long[][] c = calcC(n);
			long[] v = new long[n];
			Map<Long, Integer> map = new HashMap<>();
			IntStream.range(0, n).forEach(i -> {
				v[i] = scanner.nextLong();
				map.put(v[i], map.getOrDefault(v[i], 0) + 1);
			});
			Arrays.sort(v);
			System.out.println(IntStream.rangeClosed(1, a).mapToLong(i -> v[n - i]).average().getAsDouble());
			if (v[n - 1] == v[n - a]) {
				Integer count = map.get(v[n - 1]);
				int max = Math.min(count, b);
				System.out.println(IntStream.rangeClosed(a, max).mapToLong(i -> c[count][i]).sum());
			} else {
				long[] keys = map.keySet().stream().sorted((l1, l2) -> l2.compareTo(l1)).mapToLong(Long::longValue)
						.toArray();
				Map<Long, Integer> countMap = new HashMap<>();
				int count = 0;
				countMap.put(keys[0], count);
				for (int i = 1; i < keys.length; i++) {
					countMap.put(keys[i], count += map.get(keys[i - 1]));
				}
				System.out.println(c[map.get(v[n - a])][a - countMap.get(v[n - a])]);
			}
		}
	}

	/**
	 * パスカルの三角形を使ってm_C_nを計算
	 * 
	 * @param n 整数の最大値
	 * @return m_C_n の配列，c[m][n]はm_C_n
	 */
	private static long[][] calcC(int n) {
		long[][] c = new long[n + 1][n + 1];
		c[0][0] = 1;
		IntStream.rangeClosed(1, n).forEach(i -> IntStream.rangeClosed(0, i).forEach(j -> {
			// c[i][j] = c[i-1][j-1]+c[i-1][j]
			if ((i > 0) && (j > 0)) {
				c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
			} else {
				c[i][j] = 1;
			}
		}));
		return c;
	}

	@SuppressWarnings("unused")
	private static void print(long[][] c) {
		int n = c.length;
		IntStream.range(0, n).mapToObj(i -> IntStream.rangeClosed(0, i).mapToObj(j -> String.valueOf(c[i][j]))
				.collect(Collectors.joining(","))).forEach(System.out::println);
	}
}